prove that

modulus of sin(nx)<= n* modulus of sin(x)

for n being a natural number.

Sketch the graphs of both the fucntions ,

n*mod(Sin(x)) is a curve with amplitude n and period 2pi

mod(Sin(nx)) is a curve with amplitude 1 and period 2pi/n

Hi Aritra,

We can plot a graph and see that the above inequality is true.

But to actually plot the graph we need to do some mathematical analysis.

We can prove the direct inequality mathematically.

Consider |sin(nx)/sinx| = |sin[(n-1)x+x]/sinx| = |cosx + sin(n-1)xcosx/sinx|

Now write sin(n-1)x as sin[(n-2)x+x] and expand again.

We will have |cos(n-1)x + cosx*cos(n-2)x + cos^2 x*sin(n-2)x/sinx|....

We keep on doing this till we have the summation of n terms within the modulus.

Clearly each term is a trignometric term in the form of either sin/cos, and hence each term is less than 1.

Hence the summation of the n terms will be less than n.

So clearly |sin(nx)/sinx| ≤ |n| = n.

And hence |sin(nx)| ≤ n|sinx|.

Hope this helps.

All the best.

Regards,

Ashwin (IIT MadraS).